hdu 1714 RedField

RedField

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 639    Accepted Submission(s): 213

Problem Description

As the graph you see below, we named the red part "RedField".The read
part is the intersection of an ellipse and a triangle.Now, 8600's not
good at math, so he wants you to help him calculate the area of the
"RedField".

 

Input

The first line of input contains n, the number of test cases. n lines
follow.Each test case contains four numbers a, b, x, y(0 < b <= a
<= x),and a is the lenth of OA, b is the lenth of OB,and x, y
representing the coordinate of the point P.

 

Output

For each test case, output the area of the "RedField",accurate up to 2 decimal places.

 

Sample Input

1
1.00 1.00 2.00 2.00

 

Sample Output

0.39

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

非常简单的高数题！

1.先求出交点！

2.再对交点右边的含有抛物线的部分积分，再加上三角形就好了！

有一个要注意的地方：就是y可以为负数，即p可以在x轴的下方。（这里我WA了3次，哎真的没有仔细考虑啊！！）

1. #include<iostream>
2. #include<cstdio>
3. #include<cmath>
4. using namespace std;
5. int main()
6. {
7. int n;
8. cin>>n;
9. while(n--)
10. {
11. double A,B,k,j_x,j_y,ans;
12. double p_x,p_y;
13. cin>>A>>B>>p_x>>p_y;
14. p_y = p_y>0?p_y:-p_y;
15. k=p_y/p_x;
16. double tt=(A*A*B*B)/(B*B+A*A*k*k);
17. j_x=sqrt(tt);
18. j_y=k*j_x;
19. double t1=asin(1.0);
20. double t2=asin(j_x/A);
21. ans=A*B*(  ( t1+sin(2*t1)/2.0) -(t2+sin(2*t2)/2.0) )/2+j_x*j_y/2;
22. printf("%.2lf\n",ans);
23. }
24. return 0;
25. }